Date: Address Contents 0x10010004 0x213FED1E 0x1001000 Ox1 122334F Register Cont
ID: 3705005 • Letter: D
Question
Date: Address Contents 0x10010004 0x213FED1E 0x1001000 Ox1 122334F Register Contents 0x10010000 0x98272ACA $21 | 822 | 0x1001000C | | 0x10010008 | 0x219DEFAD Consider the following information above Please give the full 32-bit value inside the register in hexadecimal.: lb $21, -3(822) 1. (5 Points) What is the contents of register 21 after executing the following instruction. 2. (5 Points) What is the contents of register 21 after executing the following instruction. Please give the full 32-bit value inside the register in hexadecimal. lw $21, -8(822)Explanation / Answer
1.
lb $21 , -3($22) //load byte at address ($22-3) into $21
Content at $22 -3 i.e 0x10010009 is 9D
Content of $21 becomes 0x0000009D
2.
lw $21 , -8($22) //Load contents of address ($22 - 8) into register $21
($22 - 8) = 0x1001000C - 0x10010004
= 0x10010004
content at address 0x10010004 is stored in $21 i.e. content of $21 = 0x213FED1E
Register Contents $21 $22 0x1001000C